Any set $X$ with a linear order has a uniquely associated order topology generated by the open intervals. That makes it into a linearly ordered topological space (LOTS).
It is also a standard result that any countable linear order is isomorphic as a linear order to a subset of $\Bbb Q$ (uses the fact that $(\Bbb Q, <)$ is a dense linear order). See for example here.
Now in general there are many ways a linear order can be embedded into $\Bbb Q$. For example consider these two subsets:
$$A_1=\{0\}\cup\{1/n:n=1,2,\ldots\}$$
$$A_2=\{-1\}\cup\{1/n:n=1,2,\ldots\}$$
They are isomorphic to each other as linear order, and hence have the same intrinsic LOTS topology. But as topological subspaces of $\Bbb Q$ they are very different. The first one is compact, and the second one has the discrete topology, but only the first one has its intrinsic LOTS topology match its topology as a subspace of $\Bbb Q$. So an embedding that makes the subspace and order topologies match is special in a way.
Here is my question:
Given a countable set $X$ with a linear order, is it possible to embed it into $\Bbb Q$ by a linear order isomorphism so that the order topology on $X$ matches the subspace topology from $\Bbb Q$?
Anything you want to add explaining the relationship between the two topologies would be very interesting. For example one topology always stronger than the other. And are there some cases for the linear order $X$ where the two topologies always coincide independently of a linear order embedding?
Best Answer
I believe the answer is yes.
Call a subset $A\subseteq \mathbf Q$ well-embedded if the intrinsic and subspace topologies coincide. The basic idea is that the failure of well-embededness is caused by a set of "holes" between elements of $A$ and monotone sequences convergent to them in $A$, and we can upgrade any subset of $\mathbf Q$ to a well-embedded one by "collapsing" the corresponding holes.
Note that the topologies coincide exactly when every point has the same neighbourhoods in both topologies. It is not hard to see that $A$ is well-embedded exactly when for each $a\in A$, if $a$ is neither minimal nor a successor (in $A$), then for every $q<a$ there is some $a'\in A$ such that $q\leq a'<a$, and if it is neither a predecessor nor maximal, then for every $q>a$, there is some $a'$ with $a<a'\leq q$.
Conversely, $A$ is not well-embedded exactly when there is some half-open interval $(a,q]$ such that $a$ is not a predecessor and not maximal in $A$ and $(a,q]\cap A$ is empty, or some half-open interval $[q,a)$ with the corresponding property. Let us call such an interval a bad interval of $A$. Note that every element of a bad interval is also the (closed) endpoint of a bad interval.
Fix any $A\subseteq\mathbf Q$. Write $L=\mathbf Q\setminus \bigcup_I I$, where $I$ ranges over bad intervals. Note that $A\subseteq L$. I claim that $L$ is dense, and so $L'=\mathbf Q+(\mathbf Q\setminus \bigcup_I I)+\mathbf Q$ is dense without endpoints.
Indeed, suppose towards contradiction that $q_1< q_2\in L$ are such that $(q_1,q_2)\cap L$ is empty. Take some $q\in (q_1,q_2)$. Then $q$ is the closed endpoint of a bad interval $(a,q]$ or $[q,a)$. Suppose the former holds (the other case is analogous). Then we cannot have $a<q_1$ (because then $q_1\in (a,q]$) and we cannot have $a>q_1$ (because then $a\in L$ and $q_1<a<q<q_2$, so $a\in L\cap (q_1,q_2)$), so we have $a=q_1$. Since $(q_1,q_2)\cap A=\emptyset$ and $a=q_1$ is not a predecessor in $A$, $q_2\notin A$. But then $(q_1,q_2]$ is a bad interval, a contradiction.
Note that $L'\cong \mathbf Q$ (because both are countable dense without endpoints). I claim that $A$ is well-embedded in $L'$. Indeed, if $a\in A$ is neither minimal nor a successor and $q<a$, then there is some $q'\in L$ with $q\leq q'<a$, and since the interval $[q',a)$ (in $\mathbf Q$) is not bad, there is some $a'\in [q',a)$, whence $q\leq a'<a$. The other case to consider is analogous, and this completes the proof.
(In fact, I'm pretty sure that if you work just a little bit harder, you can show that a) $L$ already has no endpoints, and b), for any $A\subseteq \mathbf Q$, there is a weakly monotone, piecewise linear $f\colon \mathbf Q\to \mathbf Q$ which is strictly monotone on $A$ and such that $f[A]$ is well-embedded.)